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2 years ago

As people exited a movie theater, they were informally surveyed about whether they enjoyed the movie or not.

Mathematics

1 answer:

Ilia_Sergeevich [38]2 years ago

4 0

A=41% B=3% hope this helps

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9. On Johnny math test, he was asked to re-write the expression –3(x + 10) by using the distributive

kirill [66]

Answer:

No, he is incorrect because he forgot to multiple the negative by the x, and only multiplied 3 by x instead of -3 by x.

Step-by-step explanation:

The correct re-write of this expression is:

-3x - 30

5 0

3 years ago

There are (4^9)5 ⋅ 4^0 books at the library. What is the total number of books at the library?

LenKa [72]

Your answer is B. You multiply 5 by the exponents and 4^0 is just 1, so 4^45•1=4^45. Hope this helped!

7 0

3 years ago

A line passes through the points

masha68 [24]

Answer:

y=-2x+7

enter the point into point slope form to find the slope and then use one of the points in point slope form and rearrange it to have y by itself

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3 years ago

Avery bought 6.5 pounds of produce. If the produce was priced at $1.88 per pound, how much change did Avery receive if she paid

BARSIC [14]

Answer:

7.78

Step-by-step explanation:

6.5 x 1.88 = 12.22

20.00 - 12.22 = 7.78

6 0

3 years ago

One household is to be selected at random from a town. The probability that the household has a cat is 0.20.2 . The proba

dimulka [17.4K]

Answer:

There is a 50% probability that the household has a dog, given that the household has a cat.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a household has a cat.

B is the probability that a household has a dog.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a household has a cat but not a dog and $A \cap B$ is the probability that a household has both a cat and a dog.

By the same logic, we have that:

$B = b + (A \cap B)$

The probability that the household has a cat or a dog is 0.5

$a + b + (A \cap B) = 0.5$

The probability that the household has a dog is 0.4

$B = 0.4$

$B = b + (A \cap B)$

$b = 0.4 - (A \cap B)$

The probability that the household has a cat is 0.2.

$A = 0.2$

$A = a + (A \cap B)$

$a = 0.2 - (A \cap B)$

$a + b + (A \cap B) = 0.5$

$0.2 - (A \cap B) + 0.4 - (A \cap B) + (A \cap B) = 0.5$

$A \cap B = 0.1$

What is the probability that the household has a dog, given that the household has a cat?

20% of the households have a cat, and 10% have both a cat and a dog. So

$P = \frac{A \cap B}{A} = {0.1}{0.2} = 0.5$

There is a 50% probability that the household has a dog, given that the household has a cat.

4 0

3 years ago